AMU Medical AMU Solved Paper-2008

  • question_answer
    The square of resultant of two equal forces is three times their product. Angle between the forces is

    A) \[\pi \]                                 

    B)  \[\frac{\pi }{2}\]

    C) \[\frac{\pi }{4}\]                                              

    D)  \[\frac{\pi }{3}\]

    Correct Answer: D

    Solution :

                     Let \[\theta \] be the angle between vectors \[\vec{P}\] and \[\vec{Q}\] whose resultant is R. Here, P = Q and \[{{R}^{2}}=3PQ=3{{P}^{2}}\] As           \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta \]                 \[\therefore \]  \[3{{P}^{2}}={{P}^{2}}+{{P}^{2}}+2{{P}^{2}}\cos \theta \] or            \[3{{P}^{2}}-2{{P}^{2}}=2{{P}^{2}}\cos \theta \] or            \[{{P}^{2}}=2{{P}^{2}}\cos \theta \] or            \[1=2\cos \theta \] \[\therefore \cos \theta =\frac{1}{2}\], thus, \[\cos \theta =cos{{60}^{o}}\] or            \[\theta ={{60}^{o}}=\frac{\pi }{3}\]


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